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Let $A$ be an abelian variety over an algebraically closed field $k$.

Let $\phi:A\to A$ be an étale isogeny (over $k$). Suppose that the set $\cup_{r\geq 0}({\rm ker}\,\phi^{\circ r})(k)$ is Zariski dense in $A$. Here $\phi^{\circ r}:=\phi\circ\dots\circ\phi$ ($r$-times).

Now let $\lambda:A\to A$ be an endomorphism and suppose that for all $r\geq 0$, we have $$\lambda(({\rm ker}\,\phi^{\circ r})(k))\subseteq ({\rm ker}\,\phi^{\circ r})(k)\,\,\,\,(*).$$

Does it follow that we have $$\phi^{-1}\circ\lambda\circ \phi=\lambda\circ c$$ in ${\rm End}(A)_{\bf Q}$ for some $c\in{\rm End}(A)_{\bf Q}$ which commutes with $\phi$? Or more generally, what relation in ${\rm End}(A)_{\bf Q}$ between $\lambda$ and $\phi$ does $(*)$ imply?

One may of course ask the same question for any isogeny (not just étale) and formulate a more general condition involving finite group schemes but the case above is the case I am interested in.

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    $\begingroup$ Couldn't you have something like $\lambda\circ\phi=- \phi\circ\lambda$ (or, more generally, automorphisms of finite order)? $\endgroup$ – ACL Mar 29 '17 at 11:47
  • $\begingroup$ @ACL: thank you. You are quite right (silly me!). In fact one could replace $-1$ by any endomorphism commuting with $\phi$. I shall modify the question accordingly but I am bit confused about what to expect. $\endgroup$ – Damian Rössler Mar 29 '17 at 12:49
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    $\begingroup$ No, not any endomorphism : equality of degrees forces automorphism. $\endgroup$ – ACL Mar 29 '17 at 17:00
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    $\begingroup$ @ACL: only if $\lambda$ is an isogeny. $\endgroup$ – Damian Rössler Mar 30 '17 at 8:42
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Remark. The condition that $\bigcup_{r \geq 0} \ker(\phi^r)(k)$ is Zariski dense is not needed. Indeed, if it is not satisfied, replace $\phi$ by $[\ell] \circ \phi$ for $\ell$ invertible in $k$. Since $([\ell] \circ \phi)^r = [\ell^r] \circ \phi^r$, the kernel contains all $\ell^r$-torsion, which is Zariski dense as $r \to \infty$.

Moreover, $[\ell] \circ \phi$ satisfies $(*)$ if $\phi$ does. Indeed, if $x \in \ker([\ell^r] \circ \phi^r)$, then $$([\ell^r] \circ \phi^r) (\lambda(x)) = \phi^r(\lambda([\ell^r]x)) = 0,$$ since $[\ell^r]x \in \ker \phi^r$ and by condition $(*)$ for $\phi$. Thus, $\lambda(x) \in \ker([\ell^r] \circ \phi^r)$, which proves condition $(*)$ for $[\ell] \circ \phi$.

Finally, $\phi^{-1} \circ \lambda \circ \phi$ is unaffected by this replacement, since scalars are central.

Negative answer to your question. Let $A = E \times E$ be the square of an elliptic curve. Let $\phi \colon (x,y) \mapsto (y,x)$ be the swap isomorphism, and let $\lambda \colon (x,y) \mapsto (x,0)$ be the projection. Condition $(*)$ is now trivially satisfied (there is no kernel).

But $\phi^{-1} \circ \lambda \circ \phi$ is now the other coordinate projection $(x,y) \mapsto (0,y)$. This is not of the form $\lambda \circ c$ for any $c$, for example since the images don't agree (even after inverting scalars). Nor is it of the form $c \circ \lambda$, because the kernels don't agree.

Is there anything we can say? The best I can do is the following: condition $(*)$ implies, by the fundamental theorem on homomorphisms¹, that there exist maps $\lambda_r \colon A \to A$ such that the diagrams $$\begin{array}{ccc}A & \stackrel\lambda\longrightarrow & A \\ {\tiny{\phi^r}}\downarrow\ \ & & \ \ \downarrow\tiny{\phi^r} \\ A & \stackrel{\lambda_r}\longrightarrow & A \end{array}$$ commute. In fact, this is equivalent to condition $(*)$: the converse follows from the diagram.

But as we saw above, the relation between $\lambda_1$ and $\lambda$ is not something simple like $\lambda_1 = \lambda \circ c$ or $\lambda_1 = c \circ \lambda$ for some $c$.


¹One should say some words about set-theoretic inclusion and scheme-theoretic inclusion in $(*)$. This should be fine because $\ker(\phi^r)$ is an étale $k$-scheme, and $k$ is algebraically closed.

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  • $\begingroup$ Right! Thank you very much. That clarifies it - one cannot say anything more than what the diagram expresses in general. $\endgroup$ – Damian Rössler Mar 30 '17 at 8:39

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